Dear Sudent ρ =A+Bx is linear density varying with xLet's take dx element of this rodSo mass dm of this element will be equal todm =density per unit length × lengthdm =A+BxdxNow centre of mass xcm =∫x dm∫dm =∫0Lx A+Bxdx∫0LA+Bxdxxcm =∫0LAx dx+∫0LBx2 dx∫0LA dx+∫0LBx dx =AL22 +BL33AL +BL22xcm = 3AL2+2BL362AL+BL22 =3A+2BL3(2A+BL)L Regards

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Dear Sudent

ρ = A + B x i s l i n e a r d e n s i t y v a r y i n g w i t h x L e t' s t a k e d x e l e m e n t o f t h i s r o d S o m a s s d m o f t h i s e l e m e n t w i l l b e e q u a l t o d m = d e n s i t y p e r u n i t l e n g t h \times l e n g t h d m = (A + B x) d x N o w c e n t r e o f m a s s x_{c m} = \frac{\int x d m}{\int d m} = \frac{\int_{0}^{L} x (A + B x) d x}{\int_{0}^{L} (A + B x) d x} x_{c m} = \frac{\int_{0}^{L} A x d x + \int_{0}^{L} B x^{2} d x}{\int_{0}^{L} A d x + \int_{0}^{L} B x d x} = \frac{\frac{A L^{2}}{2} + \frac{B L^{3}}{3}}{A L + \frac{B L^{2}}{2}} x_{c m} = \frac{\frac{3 A L^{2} + 2 B L^{3}}{6}}{\frac{2 A L + B L^{2}}{2}} = \frac{(3 A + 2 B L)}{3 (2 A + B L)} L

Regards

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